Question

Based on the following atomic mass values $-1\mathrm{H}$, 1.00782 amu; ${}^2\mathrm{H},2.01410\mathrm{amu};{}^3\mathrm{H},3.01605\mathrm{amu};{}^3\mathrm{He}$ 3.01603 amu; ${}^4$ He, 4.00260 amu- and the mass of the neutron given in the text, calculate the energy released per mole in each of the following nuclear reactions, all of which are possibilities for a controlled fusion process: (a) ${}_1^2\mathrm{H}+{}_1^3\mathrm{H}⟶{}_2^4\mathrm{He}+{}_0^1\mathrm{n}$ (b) ${}_1^2\mathrm{H}+{}_1^2\mathrm{H}⟶{}_2^3\mathrm{He}+{}_0^1\mathrm{n}$ (c) ${}_1^2\mathrm{H}+{}_2^3\mathrm{He}⟶{}_2^4\mathrm{He}+{}_1^1\mathrm{H}$

Short answer

The energy released per mole for each fusion reaction is: (a) 1.684 × 10^12 J/mol (b) 3.131 × 10^11 J/mol (c) 1.759 × 10^12 J/mol

Step-by-step solution

Calculate the mass difference for each reaction

For each reaction, we will find the mass difference between the reactants and products. The mass difference will be used to calculate the energy released per mole in the next step. (a) $$\begin{array}{rl}\mathrm{\Delta }{m}_a& =(m{(}^{2}H)+m{(}^{3}H))-(m{(}^{4}He)+m{(}^{1}n))\\ & =(2.01410+3.01605)-(4.00260+1.00866)\\ & =5.03015-5.01126\\ & =0.01889\text{ amu}\end{array}$$ (b) $$\begin{array}{rl}\mathrm{\Delta }{m}_b& =(m{(}^{2}H)+m{(}^{2}H))-(m{(}^{3}He)+m{(}^{1}n))\\ & =(2.01410+2.01410)-(3.01603+1.00866)\\ & =4.02820-4.02469\\ & =0.00351\text{ amu}\end{array}$$ (c) $$\begin{array}{rl}\mathrm{\Delta }{m}_c& =(m{(}^{2}H)+m{(}^{3}He))-(m{(}^{4}He)+m{(}^{1}H))\\ & =(2.01410+3.01603)-(4.00260+1.00782)\\ & =5.03013-5.01042\\ & =0.01971\text{ amu}\end{array}$$

Convert mass difference to energy released per mole

Now that we have the mass difference for each reaction, we can convert it to energy released per mole using the mass-energy equivalence relationship and the conversion factor from amu to kg. Mass of 1 amu = 1.66054 × 10^-27 kg Speed of light, c = 2.998 × 10^8 m/s (a) $$\begin{array}{rl}{E}_a& =\mathrm{\Delta }{m}_a\times (1.66054\times {10}^{-27}\text{ kg/amu})\times (2.998\times {10}^8\text{ m/s}{)}^2\\ & =0.01889\times 1.66054\times {10}^{-27}\text{ kg}\times (2.998\times {10}^8\text{ m/s}{)}^2\\ & =2.799\times {10}^{-12}\text{ J}\end{array}$$ (b) $$\begin{array}{rl}{E}_b& =\mathrm{\Delta }{m}_b\times (1.66054\times {10}^{-27}\text{ kg/amu})\times (2.998\times {10}^8\text{ m/s}{)}^2\\ & =0.00351\times 1.66054\times {10}^{-27}\text{ kg}\times (2.998\times {10}^8\text{ m/s}{)}^2\\ & =5.204\times {10}^{-13}\text{ J}\end{array}$$ (c) $$\begin{array}{rl}{E}_c& =\mathrm{\Delta }{m}_c\times (1.66054\times {10}^{-27}\text{ kg/amu})\times (2.998\times {10}^8\text{ m/s}{)}^2\\ & =0.01971\times 1.66054\times {10}^{-27}\text{ kg}\times (2.998\times {10}^8\text{ m/s}{)}^2\\ & =2.922\times {10}^{-12}\text{ J}\end{array}$$

Convert energy released to energy per mole

Finally, we will convert the energy released for each reaction to energy per mole by multiplying by Avogadro's number (6.022 × 10^23 mol^-1). (a) $$E_a\text{ per mole}=2.799\times {10}^{-12}\text{ J}\times 6.022\times {10}^{23}{\text{ mol}}^{-1}=1.684\times {10}^{12}\text{ J/mol}$$ (b) $$E_b\text{ per mole}=5.204\times {10}^{-13}\text{ J}\times 6.022\times {10}^{23}{\text{ mol}}^{-1}=3.131\times {10}^{11}\text{ J/mol}$$ (c) $$E_c\text{ per mole}=2.922\times {10}^{-12}\text{ J}\times 6.022\times {10}^{23}{\text{ mol}}^{-1}=1.759\times {10}^{12}\text{ J/mol}$$ So the energy released per mole for each fusion reaction is: (a) 1.684 × 10^12 J/mol (b) 3.131 × 10^11 J/mol (c) 1.759 × 10^12 J/mol

Key concepts

Atomic Mass Unit (amu)

When dealing with atoms and subatomic particles, standard units of mass like grams or kilograms are not practical due to the extremely small masses involved. Hence, scientists use the atomic mass unit (amu), a standard unit of mass that quantifies mass on an atomic or molecular scale. One amu is defined as one twelfth the mass of a carbon-12 atom and approximately equals to $1.66054\times {10}^{-27}$ kilograms. This unit allows us to compare the masses of different atoms and molecules with relative ease.

For instance, in nuclear fusion energy calculations, exact mass values of isotopes in amu become the basis to determine the mass defect—the difference in mass before and after the reaction, which correlates directly to the energy released. The process involves converting these precise amu values into energy using the mass-energy equivalence formula. Clear understanding of the amu concept is crucial to accurately perform and interpret these calculations.

Nuclear Reaction

Nuclear reactions involve changes in an atom's nucleus and often result in energy release. Unlike chemical reactions that involve only the electron shell, nuclear reactions affect the protons and neutrons within the nucleus and can convert one element into a different one. Types of nuclear reactions include fusion, fission, and radioactive decay.

In nuclear fusion, smaller nuclei combine to form a single, larger nucleus, releasing substantial amounts of energy. This process powers our Sun and other stars, and if harnessed on Earth, could potentially provide a nearly limitless energy source. In the textbook exercise, we examine nuclear fusion reactions, which are vital for understanding how stars produce energy and the potential for fusion to be a clean energy source on Earth.

Energy Calculations in Nuclear Reactions

Performing energy calculations for nuclear reactions necessitates knowing the masses of both the reactants and the products, as even minuscule mass changes correspond to large amounts of energy, due to the high speed of light squared term in the mass-energy equivalence formula.

Mass-Energy Equivalence

One of the most famous equations in physics is Albert Einstein's mass-energy equivalence formula, represented as $E=m{c}^2$. This equation is the key to understanding nuclear reactions, where 'E' stands for energy, 'm' is mass, and 'c' is the speed of light in a vacuum. The speed of light is approximately $2.998\times {10}^8$ meters per second, and because it is squared in the formula, even a tiny mass can be converted into a tremendous amount of energy.

This principle is the core behind the energy calculations in the given nuclear fusion reactions, where the mass defect (the difference in mass between reactants and products) is converted to energy. The energy released in reactions like those in a star, or a potential fusion reactor, can be quantified through these calculations. Understanding how to correctly transpose mass (in amu) to energy (in joules) using the mass-energy equivalence is crucial for accurately predicting the amount of energy that nuclear reactions can yield.

Avogadro's Number

Avogadro's number, $6.022\times {10}^{23}$, is a fundamental constant in chemistry that denotes the number of atoms, ions, or molecules in one mole of any substance. Named after the nineteenth-century scientist Amedeo Avogadro, this constant creates a bridge between the microscopic world of atoms and the macroscopic world we interact with every day.

When dealing with energy released in nuclear reactions per mole, Avogadro's number is utilized to scale up from the energy released from a single nuclear event to the molar level. This provides a practical value that chemists and physicists can use, since dealing with individual atoms or particles is not feasible. In our nuclear fusion energy calculations, we multiply the energy obtained from the mass defect by Avogadro's number to determine the energy released per mole, transforming the atomic-scale energy release into quantities that can be measured and used in larger-scale applications.